Question: Multiply the following complex numbers: $({-2-3i}) \cdot ({3-i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-2-3i}) \cdot ({3-i}) = $ $ ({-2} \cdot {3}) + ({-2} \cdot {-1}i) + ({-3}i \cdot {3}) + ({-3}i \cdot {-1}i) $ Then simplify the terms: $ (-6) + (2i) + (-9i) + (3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -6 + (2 - 9)i + 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -6 + (2 - 9)i - 3 $ The result is simplified: $ (-6 - 3) + (-7i) = -9-7i $